Factorising Quadratics Using Their Roots

Factorising Quadratics Using Their Roots

Expand (x − 2)(x − 3): x² − 3x − 2x + 6 = x² − 5x + 6. Now the other way round: the quadratic x² − 5x + 6 can be written as the product (x − 2)(x − 3). Notice that the numbers 2 and 3 are exactly the roots of the equation x² − 5x + 6 = 0.

That's no coincidence. If you know the roots, you can factorise the quadratic, and the other way round: from the factorised form you can read off the roots straight away.

Contents

The factorising rule

If the equation ax² + bx + c = 0 has roots x₁ and x₂, then

ax² + bx + c = a · (x − x₁) · (x − x₂)

The brackets (x − x₁) and (x − x₂) are the linear factors of the quadratic. You must not leave out the number a in front of the brackets.

Why does this work? The product a(x − x₁)(x − x₂) is zero exactly when x = x₁ or x = x₂, and its coefficient of x² is a. A quadratic with the same roots and the same a is the same quadratic.

Worked examples

Worked example 1

Factorise x² + x − 12.

  1. Roots of the equation x² + x − 12 = 0: D = 1 + 48 = 49, x = (−1 ± 7) / 2, so x₁ = 3, x₂ = −4.
  2. a = 1, so x² + x − 12 = (x − 3)(x + 4).

Watch the sign: the root −4 gives the factor x − (−4) = x + 4.

Check: (x − 3)(x + 4) = x² + 4x − 3x − 12 = x² + x − 12 ✓.

Worked example 2 — the number a in front of the brackets

Factorise 2x² − 2x − 12.

  1. Roots: D = 4 + 96 = 100, x = (2 ± 10) / 4, so x₁ = 3, x₂ = −2.
  2. a = 2, so 2x² − 2x − 12 = 2(x − 3)(x + 2).

Worked example 3 — a fractional root

Factorise 3x² + 4x + 1.

  1. D = 16 − 12 = 4, x = (−4 ± 2) / 6, so x₁ = −1/3, x₂ = −1.
  2. 3x² + 4x + 1 = 3(x + 1/3)(x + 1).
  3. Multiply the 3 into the first bracket: 3(x + 1/3) = 3x + 1. Result: (3x + 1)(x + 1).

Check: (3x + 1)(x + 1) = 3x² + 3x + x + 1 = 3x² + 4x + 1 ✓.

Worked example 4 — a repeated root

x² − 6x + 9 has only one root, 3 (D = 0), so x² − 6x + 9 = (x − 3)(x − 3) = (x − 3)².

When a quadratic cannot be factorised

If D < 0, the equation has no real roots and the quadratic cannot be factorised into linear factors with real numbers. For example x² + x + 1 (D = 1 − 4 = −3).

Building an equation from its roots

The rule also works the other way round: if you know the roots, you build the equation by expanding the brackets.

Worked example 5

Write an equation with roots 2 and −5 (a = 1).

(x − 2)(x + 5) = x² + 5x − 2x − 10, so x² + 3x − 10 = 0.

Worked example 6

Write an equation with roots 1/2 and 3 and with a = 2.

2(x − 1/2)(x − 3) = (2x − 1)(x − 3) = 2x² − 6x − x + 3, so 2x² − 7x + 3 = 0.

Worked example 7 — roots with surds

Write an equation with roots 2 + √3 and 2 − √3 (a = 1).

(x − 2 − √3)(x − 2 + √3) = (x − 2)² − (√3)² = x² − 4x + 4 − 3, so x² − 4x + 1 = 0.

We used the difference of two squares (A − B)(A + B) = A² − B² with A = x − 2 and B = √3.

What factorising is good for

  • Simplifying algebraic fractions. (x² − 5x + 6) / (x − 2) = (x − 2)(x − 3) / (x − 2) = x − 3 for x ≠ 2.
  • Reading off the roots quickly. From the form 2(x − 3)(x + 2) = 0 you can see the solutions 3 and −2 straight away.
  • Quadratic inequalities. It is easier to work out the sign of a product than the sign of the whole quadratic.

Common mistakes

  • Leaving out a. 2x² − 2x − 12 is not (x − 3)(x + 2): that is only x² − x − 6.
  • Signs. The root −4 gives (x + 4), not (x − 4).
  • A fraction in the bracket. 3(x + 1/3) = 3x + 1, not 3x + 3.
  • Factorising when D < 0. If the equation has no real roots, there is no factorisation into real linear factors.

In short

  • If a quadratic has roots x₁, x₂, then ax² + bx + c = a(x − x₁)(x − x₂).
  • You find the roots with the formula (or, when a = 1, by spotting two numbers that multiply to c and add to b).
  • The root −k gives the factor (x + k).
  • With a repeated root the factorised form is a(x − x₁)².
  • To build an equation from its roots, expand a(x − x₁)(x − x₂).

Try it yourself

  1. Factorise x² − 7x + 10.
  2. Factorise x² + 2x − 15.
  3. Factorise 2x² + 5x − 3.
  4. Write an equation with roots −1 and 4 (a = 1).
  5. Write an equation with roots 1 + √2 and 1 − √2 (a = 1).
Solutions:
  1. Roots 2 and 5: (x − 2)(x − 5).
  2. Roots 3 and −5: (x − 3)(x + 5).
  3. D = 25 + 24 = 49, roots 1/2 and −3: 2(x − 1/2)(x + 3) = (2x − 1)(x + 3).
  4. (x + 1)(x − 4) = x² − 3x − 4, equation x² − 3x − 4 = 0.
  5. (x − 1)² − 2 = x² − 2x − 1, equation x² − 2x − 1 = 0.

Practise

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